Question

a) A bridge is 30 ft above a canal. A motorboat going 10ft/sec passes under the bridge at the same instant a man walking 5ft/sec reaches that point. How rapidly are they seperating 3 seconds later?

B) If the man reaches the center of the bridge 5 seconds before the passes under it find the rate of which the distance between them is changing 4 seconds after the man crosses

Answer 1

If the river flows in the +y direction, and
the man is walking in the +x direction, then
the distance s between the two is
s^2 = x^2 + y^2 + 30^2
So, since x and y depend on t,
s^2 = (10t)^2 + (5t)^2 + 30^2 = 125t^2 + 900
s ds/dt = 250t
at t=3, s^2 = 30^2+15^2+30^2 = 45^2
45 ds/dt = 250*3
ds/dt = 16.67 ft/s

(B) ??? what difference does it make when the man reaches the center of the bridge? All we care about is when t=4. Since you provide no information on the width of the river, I just considered x=0 at the time when the man is directly above the boat, assuming that happened at the center of the bridge.